"미디의 정리(Midy's theorem)"의 두 판 사이의 차이

수학노트
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(사용자 3명의 중간 판 64개는 보이지 않습니다)
1번째 줄: 1번째 줄:
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">이 항목의 수학노트 원문주소</h5>
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==개요==
  
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* [['142857의 신비' 해설|'142857의 신비']]에서처럼 142857과 같은 수를 적당한 자리마다 쪼개어 더했을때 9가 많이 나타나는 현상에 대한 일반적인 이해
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** 1+8=4+5=2+7=9
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** 142 + 857=999
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** 428 + 571=999
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** 285 + 714=999
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** 857 + 142=999
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** 571 + 248=999
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** 712 + 485=999
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** 14+28+57=99
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** 42+85+71=198=2*99
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* 여기서 142857과 같은 수란 [[cyclic numbers]] 를 의미한다
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* 대부분의 성질은 [[순환군]] 을 통하여 이해할 수 있다
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*  더 구체적으로는 <math>\mathbb{Z}_p^{x}</math>에서 10^k 꼴의 원소로 생성되는 부분군과 그 coset 의 원소들의 합을 구하는 문제로 이해할 수 있다
  
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<h5>개요</h5>
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==순환마디의 길이가 2의 배수일때==
  
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;정리
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소수 p에 대하여, 분수 a/p  (<math>1\leq a \leq p-1</math>) 를 십진법 전개할 때 얻어지는 순환마디의 길이가 2n 이고, 순환마디가 <math>a_1a_2\cdots a_{n} a_{n+1}a_{n+2}\cdots a_{2n}</math> 라 하자. 다음이 성립한다
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:<math>1\leq i \leq n</math> 에 대하여, <math>a_{i} + a_{i+n}=9</math>
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또한 <math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} = 99\cdots 99</math>(n개의 9) 가 성립한다.
  
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;증명
  
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<h5>예</h5>
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분수 a/p  (<math>1\leq a \leq p-1</math>) 를 생각하자.
  
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* 142 + 857=999
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<math>g_k \equiv a10^k \pmod p</math> 를 만족시키는 <math>1\leq g_k \leq p-1</math>, <math>(k=0,1,\cdots,2n-1)</math>를 정의하자. <math>g_0=a</math> 이다.
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*  
 
  
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분수 a/p의 순환마디의 길이가 2n이면, <math>10^n \equiv -1 \pmod p</math> 가 성립하므로, <math>g_n=p-a</math> 임을 안다.
  
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<h5>재미있는 사실</h5>
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<math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n}=\frac{g_0 10^n-g_n}{p} + \frac{g_{n} 10^n-g_0}{p}=\frac{(g_0+g_n)(10^n-1)}{p}=10^n-1</math> ■
  
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* Math Overflow http://mathoverflow.net/search?q=
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==예 : 1176470588235294==
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* 네이버 지식인 http://kin.search.naver.com/search.naver?where=kin_qna&query=
 
  
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* p=17
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* 2/17 = 0.'''1176470588235294'''1176470588235294...
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* 11764705 + 88235294 = 99999999
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* 이 경우엔 위의 증명에서 <math>g_k</math> 로 쓰인 수는 2, 3, 13, 11, 8, 12, 1, 10, 15, 14, 4, 6, 9, 5, 16, 7 로 주어진다
  
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<h5>역사</h5>
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==순환마디의 길이가 3의 배수일 때==
  
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
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;정리
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* [[수학사연표 (역사)|수학사연표]]
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소수 p에 대하여, 분수 1/p 를 십진법 전개할 때 얻어지는 순환마디의 길이가 3n 이고, 순환마디가 <math>a_1a_2\cdots a_{n} a_{n+1}a_{n+2}\cdots a_{2n}a_{2n+1}a_{2n+2}\cdots a_{3n}</math> 라 하자. 다음이 성립한다.
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:<math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}= 99\cdots 99</math>
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;증명
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순환마디의 길이가 3n인 분수 1/p 를 생각하자.
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<math>g_k \equiv 10^k \pmod p</math>, <math>0\leq g_k \leq p-1</math> 라 정의하자. <math>g_0=1</math> 이다.
  
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<math>g_{2n} \equiv g_n^2 \pmod p</math>, <math>g_n^3 \equiv 1 \pmod p</math>  이므로, <math>g_0+g_n+g_{2n}\equiv 1+g_n+g_n^2=(g_n^3-1)/(g_n-1)\equiv 0 \pmod p</math> 이다.
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따라서 <math>g_0+g_n+g_{2n}=p</math> 또는 <math>g_0+g_n+g_{2n}=2p</math>가 성립한다.
  
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그러나 <math>1\leq g_k \leq p-1</math> 이므로 <math>1+g_n+g_{2n}=2p</math>일 수 없다. 따라서 <math>g_0+g_n+g_{2n}=p</math>
  
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<h5>메모</h5>
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<math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}=\frac{g_0 10^n-g_n}{p} + \frac{g_{n} 10^n-g_{2n}}{p}+\frac{g_{2n} 10^n-g_{0}}{p}=\frac{(g_0+g_n+g_{2n})(10^n-1)}{p}=10^n-1</math> ■
  
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*  일반적으로 소수 p에 대하여, 분수 1/p 를 십진법 전개할 때 얻어지는 순환마디의 길이가 3n 이라고 하자. 분수 a/p  (<math>1\leq a \leq p-1</math>) 또는 (p-a)/p  (<math>1\leq a \leq p-1</math>) 의 순환소수전개를 생각하자. 둘 중의 하나는 <math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}= 99\cdots 99</math> 다른 하나는, <math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}=2* 99\cdots 99</math> 를 만족한다
  
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==예 : 052631578947368421==
  
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<h5>관련된 항목들</h5>
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*  p=19
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**  1/19=0.'''052631578947368421'''052...
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**  52631+578947+368421=999999
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*  p=7
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**  3/7 = 0.4285714286...
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**  42+ 85+71=198
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**  4/7 = 0.5714285714
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**  57+14+28=99
  
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">수학용어번역</h5>
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==역사==
  
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* 단어사전<br>
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* 1836년 미디
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** http://www.google.com/dictionary?langpair=en|ko&q=
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
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** http://ko.wiktionary.org/wiki/
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* [[수학사 연표]]
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* 발음사전 http://www.forvo.com/search/
 
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
 
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** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
 
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* [http://www.nktech.net/science/term/term_l.jsp?l_mode=cate&s_code_cd=MA 남·북한수학용어비교]
 
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* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
 
  
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==관련된 항목들==
  
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<h5>사전 형태의 자료</h5>
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* [[합동식 (모듈로 modulo 연산)]]
  
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* http://ko.wikipedia.org/wiki/
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* [http://en.wikipedia.org/wiki/Midy%27s_theorem http://en.wikipedia.org/wiki/Midy's_theorem]
 
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* http://en.wikipedia.org/wiki/
 
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* [http://eom.springer.de/default.htm The Online Encyclopaedia of Mathematics]
 
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* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
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* [http://eqworld.ipmnet.ru/ The World of Mathematical Equations]
 
  
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==매스매티카 파일 및 계산 리소스==
  
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<h5>리뷰논문과 에세이</h5>
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* https://docs.google.com/leaf?id=0B8XXo8Tve1cxYWNhZDQwN2MtNjMyMS00ZDc2LTgzZTQtMzFmMTQxYWFmZWM0&sort=name&layout=list&num=50
  
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* [http://www.jstor.org/stable/3621748 A Curious String of Nines]<br>
 
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** Hans Liebeck,
 
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** <cite>The Mathematical Gazette</cite>, Vol. 85, No. 504 (Nov., 2001), pp. 431-438
 
  
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* [http://www.jstor.org/stable/4146879 Midy's (Nearly) Secret Theorem: An Extension after 165 Years]<br>
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** Brian D. Ginsberg, <cite>The College Mathematics Journal</cite>, Vol. 35, No. 1 (Jan., 2004), pp. 26-30
 
  
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==사전 형태의 자료==
  
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* http://ko.wikipedia.org/wiki/
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* [http://en.wikipedia.org/wiki/Midy%27s_theorem http://en.wikipedia.org/wiki/Midy's_theorem]
  
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<h5>관련논문</h5>
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*  Lewittes, Joseph. 2006. “Midy’s Theorem for Periodic Decimals”. <em>math/0605182</em> (5월 7). http://arxiv.org/abs/math/0605182<br>
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==리뷰, 에세이, 강의노트==
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* http://www.jstor.org/action/doBasicSearch?Query=
 
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* http://www.ams.org/mathscinet
 
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* http://dx.doi.org/
 
  
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* [http://www.jstor.org/stable/3621748 A Curious String of Nines]
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** Hans Liebeck,
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** <cite>The Mathematical Gazette</cite>, Vol. 85, No. 504 (Nov., 2001), pp. 431-438
  
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<h5>관련도서</h5>
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==관련논문==
  
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* 도서내검색<br>
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* Lewittes, Joseph. 2006. “Midy’s Theorem for Periodic Decimals”. <em>math/0605182</em> (5월 7). http://arxiv.org/abs/math/0605182
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** http://books.google.com/books?q=
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* A. Gupta and B. Sury, [http://www.emis.de/journals/INTEGERS/papers/f19/f19.pdf Decimal expansion of 1/p and subgroup sums], Integers: Electronic Journal Of Combinatorial Number Theory 5 (2005),
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** http://book.daum.net/search/contentSearch.do?query=
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* Brian D. Ginsberg [http://www.jstor.org/stable/4146879 Midy's (Nearly) Secret Theorem: An Extension after 165 Years], <cite>The College Mathematics Journal</cite>, Vol. 35, No. 1 (Jan., 2004), pp. 26-30
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* M. Shrader-Frechette, Complementary Rational Numbers, Math. Mag., 51 (1978) 90–98.
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* E. Midy, De quelques proprietes des nombres et des fractions decimals periodiques, Nantes, (1836), 21 pages.
  
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== 노트 ==
  
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===말뭉치===
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# In 1836, E. Midy proved that if p is a prime greater than 5, and the period of 1/p is 2 1.<ref name="ref_472101d4">[http://math.colgate.edu/~integers/h3/h3.pdf Integers: electronic journal of combinatorial number theory 7 (2007), #a03]</ref>
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# (The results of Midy and Ginsberg follow quickly from this).<ref name="ref_472101d4" />
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# E. Midy, De quelques proprietes des nombres et des fractions decimals periodiques, Nantes, (1836), 6.<ref name="ref_472101d4" />
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# This theorem, which we will examine in this paper, be- came known as Midys Theorem due to a pamphlet published by E. Midy in Nantes, France, 1836.<ref name="ref_07047442">[http://sand.truman.edu/~dgarth/klinecapstone.pdf A graphical analysis of midy’s theorem]</ref>
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# In 1836 E. Midy published at Nantes, France, a pamphlet of twenty-one pages on some topics in number theory with applications to decimals.<ref name="ref_758326cf">[http://emis.impa.br/EMIS/journals/INTEGERS/papers/h2/h2.pdf Integers: electronic journal of combinatorial number theory 7 (2007), #a02]</ref>
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# Note that (iii) explains the dierence between 1/77 where k = 3 and gcd(77, 103 which has the Midy property (1), and 1/803 where k = 4 and gcd(803, 104 which (1) fails.<ref name="ref_758326cf" />
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# Midy himself considered the case of period length 3k, but he focused on the sums ai + ai+k + ai+2k, 1 k, which do not give smooth results.<ref name="ref_758326cf" />
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# Section 2 contains the main results as we study the Midy property in a more general setting.<ref name="ref_758326cf" />
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# This is known as Midy’s theorem.<ref name="ref_1d2f78ea">[https://www.johndcook.com/blog/2018/11/19/midys-theorem/ Midy's theorem: fractions with prime denominator]</ref>
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# If k is any divisor of the period of the decimal expansion of a/p (where p is again a prime), then Midy's theorem can be generalised as follows.<ref name="ref_5a20be13">[https://en.wikipedia.org/wiki/Midy%27s_theorem Midy's theorem]</ref>
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# To prove the original Midy's theorem, take the special case where h = 2.<ref name="ref_12344c07">[https://handwiki.org/wiki/Midy%27s_theorem Midy's theorem]</ref>
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# Now this is even in number therefore 1+8 = 9 which again shows the validity of Midy’s theorem.<ref name="ref_84efc06e">[https://www.geeksforgeeks.org/midys-theorem/ Midy's theorem]</ref>
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# Given numerator and denominator, the task is to find if the resultant floating point number follows Midy’s theorem or not.<ref name="ref_84efc06e" />
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# Write( "Midy's theorem holds!" ); } else { Console.<ref name="ref_84efc06e" />
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# This is to provide the necessary machinery for the proof of Midy's theorem, as well as for completeness.<ref name="ref_b7c9355f">[https://digitalcommons.unl.edu/mathfacpub/48/ A THEOREM ON REPEATING DECIMALS]</ref>
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# In Extended Midy’s Theorem, the repeating portion is divided into m digits, then their sum is a multiple of 10m - 1.<ref name="ref_cedb9ba6">[https://www.tutorialspoint.com/extended-midy-s-theorem-in-cplusplus Extended Midy's theorem in C++]</ref>
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# 2352 9411 7647 Extended Midy's theorem holds!<ref name="ref_cedb9ba6" />
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# We then use our method to give an elementary proof of Midy’s theorem on repeating decimals.<ref name="ref_35609a5a">[https://www.semanticscholar.org/paper/A-SIMPLE-PROOF-OF-MIDY%27S-THEOREM-Hamarsheh-Jaradat/edbcb21bb9dd4e861e5512f819a0702c14c51791 [PDF] A SIMPLE PROOF OF MIDY'S THEOREM]</ref>
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# In Extended Midy’s theorem if we divide the repeating portion of a/p into m digits, then their sum is a multiple of 10m -1.<ref name="ref_c9954aeb">[https://www.geeksforgeeks.org/extended-midys-theorem/ Extended Midy's theorem]</ref>
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# Write( "Denominator is not prime, " + "thus Extended Midy's theorem " + "is not applicable" ); return ; } int l = str.<ref name="ref_c9954aeb" />
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# Write( "Extended Midy's " + "theorem holds!" ); else Console.<ref name="ref_c9954aeb" />
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# Write( "Extended Midy's " + "theorem doesn't hold!" ); } else if (l % 2 != 0) { Console.<ref name="ref_c9954aeb" />
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# The 2-block property for primes is known as Midy’s theorem (1836).<ref name="ref_8162793e">[https://www.jstor.org/stable/10.4169/002557010x479974 Repeating Decimals: A Period Piece on JSTOR]</ref>
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===소스===
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<references />
  
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<h5>링크</h5>
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== 메타데이터 ==
  
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* [http://www.ams.org/news/math-in-the-media/mathdigest-index Summaries of Media Coverage of Math]
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===위키데이터===
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* 구글 블로그 검색<br>
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* ID :  [https://www.wikidata.org/wiki/Q856158 Q856158]
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** http://blogsearch.google.com/blogsearch?q=
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===Spacy 패턴 목록===
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* [{'LOWER': 'midy'}]
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* [{'LOWER': 'midy'}, {'LOWER': "'s"}, {'LEMMA': 'theorem'}]

2021년 2월 22일 (월) 23:25 기준 최신판

개요

  • '142857의 신비'에서처럼 142857과 같은 수를 적당한 자리마다 쪼개어 더했을때 9가 많이 나타나는 현상에 대한 일반적인 이해
    • 1+8=4+5=2+7=9
    • 142 + 857=999
    • 428 + 571=999
    • 285 + 714=999
    • 857 + 142=999
    • 571 + 248=999
    • 712 + 485=999
    • 14+28+57=99
    • 42+85+71=198=2*99
  • 여기서 142857과 같은 수란 cyclic numbers 를 의미한다
  • 대부분의 성질은 순환군 을 통하여 이해할 수 있다
  • 더 구체적으로는 \(\mathbb{Z}_p^{x}\)에서 10^k 꼴의 원소로 생성되는 부분군과 그 coset 의 원소들의 합을 구하는 문제로 이해할 수 있다


순환마디의 길이가 2의 배수일때

정리

소수 p에 대하여, 분수 a/p (\(1\leq a \leq p-1\)) 를 십진법 전개할 때 얻어지는 순환마디의 길이가 2n 이고, 순환마디가 \(a_1a_2\cdots a_{n} a_{n+1}a_{n+2}\cdots a_{2n}\) 라 하자. 다음이 성립한다 \[1\leq i \leq n\] 에 대하여, \(a_{i} + a_{i+n}=9\) 또한 \(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} = 99\cdots 99\)(n개의 9) 가 성립한다.

증명

분수 a/p (\(1\leq a \leq p-1\)) 를 생각하자.

\(g_k \equiv a10^k \pmod p\) 를 만족시키는 \(1\leq g_k \leq p-1\), \((k=0,1,\cdots,2n-1)\)를 정의하자. \(g_0=a\) 이다.

분수 a/p의 순환마디의 길이가 2n이면, \(10^n \equiv -1 \pmod p\) 가 성립하므로, \(g_n=p-a\) 임을 안다.

\(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n}=\frac{g_0 10^n-g_n}{p} + \frac{g_{n} 10^n-g_0}{p}=\frac{(g_0+g_n)(10^n-1)}{p}=10^n-1\) ■


예 : 1176470588235294

  • p=17
  • 2/17 = 0.11764705882352941176470588235294...
  • 11764705 + 88235294 = 99999999
  • 이 경우엔 위의 증명에서 \(g_k\) 로 쓰인 수는 2, 3, 13, 11, 8, 12, 1, 10, 15, 14, 4, 6, 9, 5, 16, 7 로 주어진다



순환마디의 길이가 3의 배수일 때

정리

소수 p에 대하여, 분수 1/p 를 십진법 전개할 때 얻어지는 순환마디의 길이가 3n 이고, 순환마디가 \(a_1a_2\cdots a_{n} a_{n+1}a_{n+2}\cdots a_{2n}a_{2n+1}a_{2n+2}\cdots a_{3n}\) 라 하자. 다음이 성립한다. \[a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}= 99\cdots 99\]


증명

순환마디의 길이가 3n인 분수 1/p 를 생각하자.

\(g_k \equiv 10^k \pmod p\), \(0\leq g_k \leq p-1\) 라 정의하자. \(g_0=1\) 이다.

\(g_{2n} \equiv g_n^2 \pmod p\), \(g_n^3 \equiv 1 \pmod p\) 이므로, \(g_0+g_n+g_{2n}\equiv 1+g_n+g_n^2=(g_n^3-1)/(g_n-1)\equiv 0 \pmod p\) 이다.

따라서 \(g_0+g_n+g_{2n}=p\) 또는 \(g_0+g_n+g_{2n}=2p\)가 성립한다.

그러나 \(1\leq g_k \leq p-1\) 이므로 \(1+g_n+g_{2n}=2p\)일 수 없다. 따라서 \(g_0+g_n+g_{2n}=p\)

\(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}=\frac{g_0 10^n-g_n}{p} + \frac{g_{n} 10^n-g_{2n}}{p}+\frac{g_{2n} 10^n-g_{0}}{p}=\frac{(g_0+g_n+g_{2n})(10^n-1)}{p}=10^n-1\) ■

  • 일반적으로 소수 p에 대하여, 분수 1/p 를 십진법 전개할 때 얻어지는 순환마디의 길이가 3n 이라고 하자. 분수 a/p (\(1\leq a \leq p-1\)) 또는 (p-a)/p (\(1\leq a \leq p-1\)) 의 순환소수전개를 생각하자. 둘 중의 하나는 \(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}= 99\cdots 99\) 다른 하나는, \(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}=2* 99\cdots 99\) 를 만족한다

예 : 052631578947368421

  • p=19
    • 1/19=0.052631578947368421052...
    • 52631+578947+368421=999999
  • p=7
    • 3/7 = 0.4285714286...
    • 42+ 85+71=198
    • 4/7 = 0.5714285714
    • 57+14+28=99



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말뭉치

  1. In 1836, E. Midy proved that if p is a prime greater than 5, and the period of 1/p is 2 1.[1]
  2. (The results of Midy and Ginsberg follow quickly from this).[1]
  3. E. Midy, De quelques proprietes des nombres et des fractions decimals periodiques, Nantes, (1836), 6.[1]
  4. This theorem, which we will examine in this paper, be- came known as Midys Theorem due to a pamphlet published by E. Midy in Nantes, France, 1836.[2]
  5. In 1836 E. Midy published at Nantes, France, a pamphlet of twenty-one pages on some topics in number theory with applications to decimals.[3]
  6. Note that (iii) explains the dierence between 1/77 where k = 3 and gcd(77, 103 which has the Midy property (1), and 1/803 where k = 4 and gcd(803, 104 which (1) fails.[3]
  7. Midy himself considered the case of period length 3k, but he focused on the sums ai + ai+k + ai+2k, 1 k, which do not give smooth results.[3]
  8. Section 2 contains the main results as we study the Midy property in a more general setting.[3]
  9. This is known as Midy’s theorem.[4]
  10. If k is any divisor of the period of the decimal expansion of a/p (where p is again a prime), then Midy's theorem can be generalised as follows.[5]
  11. To prove the original Midy's theorem, take the special case where h = 2.[6]
  12. Now this is even in number therefore 1+8 = 9 which again shows the validity of Midy’s theorem.[7]
  13. Given numerator and denominator, the task is to find if the resultant floating point number follows Midy’s theorem or not.[7]
  14. Write( "Midy's theorem holds!" ); } else { Console.[7]
  15. This is to provide the necessary machinery for the proof of Midy's theorem, as well as for completeness.[8]
  16. In Extended Midy’s Theorem, the repeating portion is divided into m digits, then their sum is a multiple of 10m - 1.[9]
  17. 2352 9411 7647 Extended Midy's theorem holds![9]
  18. We then use our method to give an elementary proof of Midy’s theorem on repeating decimals.[10]
  19. In Extended Midy’s theorem if we divide the repeating portion of a/p into m digits, then their sum is a multiple of 10m -1.[11]
  20. Write( "Denominator is not prime, " + "thus Extended Midy's theorem " + "is not applicable" ); return ; } int l = str.[11]
  21. Write( "Extended Midy's " + "theorem holds!" ); else Console.[11]
  22. Write( "Extended Midy's " + "theorem doesn't hold!" ); } else if (l % 2 != 0) { Console.[11]
  23. The 2-block property for primes is known as Midy’s theorem (1836).[12]

소스

  1. ↑ 1.0 1.1 1.2 Integers: electronic journal of combinatorial number theory 7 (2007), #a03
  2. ↑ A graphical analysis of midy’s theorem
  3. ↑ 3.0 3.1 3.2 3.3 Integers: electronic journal of combinatorial number theory 7 (2007), #a02
  4. ↑ Midy's theorem: fractions with prime denominator
  5. ↑ Midy's theorem
  6. ↑ Midy's theorem
  7. ↑ 7.0 7.1 7.2 Midy's theorem
  8. ↑ A THEOREM ON REPEATING DECIMALS
  9. ↑ 9.0 9.1 Extended Midy's theorem in C++
  10. ↑ [PDF A SIMPLE PROOF OF MIDY'S THEOREM]
  11. ↑ 11.0 11.1 11.2 11.3 Extended Midy's theorem
  12. ↑ Repeating Decimals: A Period Piece on JSTOR

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